Two simple pendulums of length 0.5 m and 20 m, respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed oscillations.

Options

A. 5
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to find the time periods of two pendulums with lengths of 0.5 m and 20 m, and determine when they will be in phase again after the shorter pendulum has completed its oscillations.

### Step-by-Step Solution:

1. **Identify the lengths of the pendulums:**
- Let \( L_1 = 0.5 \, \text{m} \) (shorter pendulum)
- Let \( L_2 = 20 \, \text{m} \) (longer pendulum)

2. **Use the formula for the time period of a simple pendulum:**
The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \), but we can use \( 10 \, \text{m/s}^2 \) for simplicity).

3. **Calculate the time period of the shorter pendulum \( T_1 \):**
\[ T_1 = 2\pi \sqrt{\frac{L_1}{g}} = 2\pi \sqrt{\frac{0.5}{10}} = 2\pi \sqrt{0.05} = 2\pi \cdot 0.2236 \approx 1.41 \, \text{s} \]

4. **Calculate the time period of the longer pendulum \( T_2 \):**
\[ T_2 = 2\pi \sqrt{\frac{L_2}{g}} = 2\pi \sqrt{\frac{20}{10}} = 2\pi \sqrt{2} \approx 2\pi \cdot 1.414 \approx 8.88 \, \text{s} \]

5. **Determine the ratio of the time periods:**
To find when the two pendulums will be in phase again, we need to find the least common multiple (LCM) of their periods. The ratio of their periods is:
\[ \frac{T_2}{T_1} = \frac{8.88}{1.41} \approx 6.29 \]
This indicates that the longer pendulum takes about 6.29 times longer than the shorter pendulum to complete one oscillation.

6. **Find the number of oscillations:**
The shorter pendulum will complete one full oscillation in \( T_1 \) seconds. To find out how many oscillations the longer pendulum completes in that time, we can calculate:
\[ \text{Number of oscillations of } T_2 = \frac{T_1}{T_2} = \frac{1.41}{8.88} \approx 0.158 \]
This means that the longer pendulum will complete approximately \( 0.158 \) of its oscillation when the shorter pendulum completes one full oscillation.

7. **Conclusion:**
The shorter pendulum will complete its oscillation while the longer pendulum is still oscillating. They will be back in phase when the shorter pendulum completes its oscillation, which is when the longer pendulum has completed \( 2 \) full oscillations.

### Final Answer:
The shorter pendulum will complete its oscillations while the longer pendulum will have completed \( 2 \) oscillations.